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Talk:Subcubic graph number
Where the different values of SCG(n)? And what level of growth in this function? In list of googologisms it is considered as uncomputable number. Is this right? Ikosarakt1 (talk) 11:11, November 9, 2012 (UTC) Even SCG(0) seems hard to compute. We can start with a graph with 1 vertex, which can contain 1 loop. This means there are no more loops in any of the remaining graphs. The second graph can have 2 vertices, with three edges connecting them. The third graph can have 3 vertices, with two edges connecting every pair. The fourth graph can have 4 vertices a,b,c,d, with ab, bc, and cd connected by two edges each, and the remaining three pairs connected by one edge each. It seems that this can go on for a very long time. So even SCG(0) might be very large. The growth rate of SCG(n) is at least the level of ψΩ1 (Ωω) in the fast-growing hierarchy. One of these days I will write a blog post explaining the fast-growing hierarchy up to these large ordinals. I'm pretty sure that SCG(13) will be greater than meameamealokkapoowa oompa, and Chris Bird comes to the same conclusion in his Bowers' Named Numbers article. "Uncomputable number" is a rather vague term, but it is certainly true that SCG(n) is a computable function. You can write a program that runs through all possible acceptable sequences of graphs and keeps track of the longest one.Deedlit11 (talk) 17:31, November 16, 2012 (UTC) Thank you that have explained it to me. I'll put it in the appropriate place. Ikosarakt1 (talk) 09:10, November 17, 2012 (UTC) Whoops, I messed up the SCG(0) analysis. I'm now quite certain that SCG(0) = 6. The first graph is a single vertex with a loop; this prevents all loops or cycles in future graphs. The second graph is two vertices with a single edge between them. This prevents all edges of any kind in future graphs. So the remaining graphs are all edgeless graphs. So the last four graphs are graphs with 3, 2, 1, and 0 vertices with no edges between them. So SCG(0) = 6, or 5 if the empty graph is not allowed. SCG(1), on the other hand, is provably very large. We can construct a sequence of graphs (for convenience I will start with G2, so that Gn has n vertices.) G2 = 2 vertices with three edges between them. G3 = vertices a, b, and c with two edges between a and b and one edge between b and c. G4 = two pairs of vertices each with two edges between them. G5 = a 5-cycle, i.e. the graph of a pentagon. G6 = a four-cycle (square) plus two vertices with an edge between them. G7 = a square plus three vertices each with a loop. G8 = a square plus two looped vertices and two unlooped vertices. G9 = a square plus two looped vertices and one unlooped vertex. G10 = a square plus two looped vertices. G11 = a square plus one looped vertex and six unlooped vertices. G12 = a square plus one looped vertex and five unlooped vertices. ... G17 = a square plus one looped vertex. G18 = a square plus 14 unlooped vertices. G19 = a square plus 13 unlooped vertices. ... G32 = a square. We can then define G33 to be a triangle plus a 30-vertex cubic tree. We can then continue with a triangle plus a very long sequence of cubic forests. Let forest(n) be the length of the longest sequence of cubic forests such that the ith forest has at most n + i vertices. Then we can continue our above sequence with a sequence of length forest(29) consisting of a triangle plus a cycleless cubic forest. We can then continue with a sequence of length forest(forest(29)) consisthing of 2-cycle plus a cyclelss cubic forest. Then we construct a sequence of length forest(forest(forest(29))) consisting for a looped vertex plus a cycleless cubic forest. We finish with a sequence of length forest(forest(forest(forest(29)))) consisting of cycleless cubic forests.' I believe the growth rate of forest(n) is about epsilon_0 in the fast-growing hierarchy, SCG(1) > forest(forest(forest(forest(29)))) is very large, clearly indeterminable.Deedlit11 (talk) 03:08, November 23, 2012 (UTC) According to you, SCG(1) is larger than goppatothplex, but it is a mere lower bound... Ikosarakt1 (talk) 12:15, November 26, 2012 (UTC) I just had an idea of how this sequence can be made a lot longer - if I'm not mistaken non-simple subcubic graphs can have loop on leaf vertex (if they can't, you can stop reading here). Define G'3 as Deedlit's G3 with loop around vertex c. G'4 is same as G3 and G'5 is same as G4. Then G'6 is hexagon graph. G'7 will be pentagon with two vertices with loops and edge between them. G'9 will be just G6 (I haven't made any analysis, so G'8 and G'9 just delete leaf loops, for simplicity). Continuing in same manner as Deedlit we'll at some point reach alone pentagon. Say it happens at G'x. Using cubic trees we can now reach forest^5(x-4). We can also insert stages between triangle plus a cycleless forest and 2-cycle plus a cycleless forest by using 2 vertices with loops around them, and then 2 vertices from which only one has loop. This creates forest^7(x-4) graphs long sequence. And for the end another lengthening - suppose that after using forests with 2-cycles we make forests and 3 looped vertices, then 2 looped vertices, then one. With some optimization we can create sequences reaching forest^(forest(x-4))(x-4) or even ones with function exponent having function exponent! LittlePeng9 (talk) 18:02, February 5, 2013 (UTC) Yeah, I wasn't trying to maximize the sequence because it was taking too much explaining to do. I would be interested to see how high you can go! Deedlit11 (talk) 04:55, March 7, 2013 (UTC) I created whole blog post defining and explaining 3 functions I later used to bound SCG(1) from below. Can you come up with better result? LittlePeng9 (talk) 20:24, March 10, 2013 (UTC) Rayo's number How does SCG(13) compare to Rayo's number? FB100Z • talk • 02:52, January 2, 2013 (UTC) Rayo's function grows as fast as \(f_{\omega_{1}^{CK} \times \omega}(n)\) (I found it at cp4space). Busy beaver function is only \(f_{\omega_{1}^{CK}}(n)\). BB(n) is yet uncomputable, and Ra(n) all the more so. SCG(n), however, is computable, so Rayo's number should be larger. Ikosarakt1 (talk) 16:49, January 19, 2013 (UTC) Look on this CP4space again. Together with A.P.Goucher we concluded that Rayo's function strength is at level of \(f_{\omega_{\omega}^{CK}}(n)\) (ordinal in there is smallest limit of admissibles, obviously still countable) LittlePeng9 (talk) 16:18, January 28, 2013 (UTC) Actual strength of SCG function Deedlit states that strength of SCG function is at least \(\psi_{\Omega_1}(\Omega_\omega)\). I guess he says so, because it is provably strength of Robertson-Seymour theorem. But this theorem is saying about well-quasi-ordering set of ''all ''graphs. Subcubic graphs make only tiny fraction of whole. I think strength of SCG function is strictly smaller than this ordinal. What do you think about that? On the other hand, at the end of Wikipedia article on graph minor theorem there is finite form concerning every possible graph with graph minorship relation. Function based on this finite form has at least \(\psi_{\Omega_1}(\Omega_\omega)\) as corresponding ordinal, and I'm almost sure I can prove that LittlePeng9 (talk) 20:42, February 13, 2013 (UTC) Actually, I wasn't basing the strength of the SCG function on the Robertson-Seymour theorem, but rather on Friedman's statements in his FOM postings. In particular, he statest that the fact that finite subcubic graphs are WQO under homeomorphic embeddability is not provable in the theory Pi-1-1-CA_0. This implies that SCG(n) is not a provably recursive function of Pi-1-1-CA_0, and the proof-theoretic ordinal of pi-1-1-CA_0 is \(\psi_{\Omega_1}(\Omega_\omega)\). In fact Friedman states that even the fact that finite planar subcubic graphs (and presumably finite planar simple subcubic graphs) are WQO under homeomorphic emeddability is not provable in Pi-1-1-CA_0, so PSCG(n) and PSSCG(n) won't be provably recursive in Pi-1-1-CA_0. Now, we may presume that a function is provably recursive in Pi-1-1-CA_0 if and only if it is elementary recursive in some function \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\). So a function not provably recursive in Pi-1-1-CA_0 will not be elementary recursive in any \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\). However, this does not mean that the function will dominate all \(F_{\alpha}\) for \( \alpha < \psi_{\Omega_1}(\Omega_\omega)\); for example, it could alternate between large values and 0. (Or, for an example of an increasing function, it could alternate between periods of growing very fast with growing very slowly, with the periods getting longer and longer.) So perhaps we cannot state with certainty that the previous functions grow at a certain rate. However, it seems highly unlikely to me that any natural function will have such strange behavior; so I am almost certain that PSSCG(n) et al grow at \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) or faster. Also, Friedman states as a theorem that SCG(13) is larger than the halting time of any Turing machine starting from the blank tape that can be proven to halt in at most \(2 \uparrow \uparrow 2000 \) symbols, based on the fact that \(SCG(13) \ge PSCG(2 \uparrow \uparrow 2000) \). So somehow he knows that \(PSCG(n) > f(n) \), where f(n) is the halting time of the longest running Turing machine starting from the blank tape that can be proven to halt in at most n symbols. I think it is clear that the latter grows at the rate of \(\psi_{\Omega_1}(\Omega_\omega)\) or higher. How Friedman knows this, I don't know, but I will trust him. Deedlit11 (talk) 04:50, March 7, 2013 (UTC) I wonder how the values of SCG(n) compare to numbers derived from the Buchholz hydra game? --Ixfd64 (talk) 18:35, March 7, 2013 (UTC) Good question! If we let BHydra(n) be the time it takes to slay a Buchholz hydra consisting of an (n+2)-path labelled with n n's (the first two vertices are required to be labelled with + and 0), then both SCG(n) and BHydra(n) are about at the level of \(\psi_{\Omega_1}(\Omega_\omega)\). BHydra(n) will in fact be roughly equivalent to \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) , as the structure of Buchholz Hydras with finite ordinal labels mimic ordinal notations up to \(\psi_{\Omega_1}(\Omega_\omega)\), and the act of slaying the hydras mimics taking fundamental sequences. On the other hand, from Friedman's work we can infer that \(PSCG(n) \ge F_{\psi_{\Omega_1}(\Omega_\omega)} \), and SCG(n) grows faster than PSCG(n) (for example \(SCG(13) > PSCG(2\uparrow\uparrow 2000) \). So I believe SCG(n) outpaces BHydra(n). However, if we label Buccholz hydras with infinite ordinals, than I bet that the numbers derived from Buchholz hydras will grow faster. Deedlit11 (talk) 00:33, March 8, 2013 (UTC) From Buchholz's work we know that for every hydra we can prove its termination in \(\Pi^1_1-CA_0\), but we can't do this uniformly (otherwise, Theorem III from this paper could be disproven by induction on number of labels). So we know that function F(n) related to (n+2)-path labelled with n ω's is unprovably recursive in \(\Pi^1_1-CA_0\), so it almost certainly outbounds \(F_{\psi_{\Omega_1}(\Omega_\omega)} \) LittlePeng9 (talk) 10:25, March 9, 2013 (UTC) 2^1000 I have heard about a^b as a notation for \(a \uparrow\uparrow b\). Is it true? Ikosarakt1 (talk) 17:38, March 1, 2013 (UTC) :Yes, really. Harvey Friedman wrote: "Here 2 ^ 1000 is an exponential stack of 2's of height 1000" Here's the link: http://www.cs.nyu.edu/pipermail/fom/2006-March/010244.htmlKonkhra (talk) 09:07, March 2, 2013 (UTC) Informal explanation? Is there a way to explain SCG less esoterically, in a way similar to TREE sequence#Explanation? It'd be nice if we could find a way to encode subcubic graphs in strings so that the concept of a graph minor is really clear. FB100Z • talk • 19:12, March 17, 2013 (UTC) I don't think so - I think best we can do is explain what graph and topological minors are, and how it connects to homeomorphic embeddability. LittlePeng9 (talk) 11:00, March 18, 2013 (UTC) I don't know of any natural interpretation of graphs as strings, other than to list out the adjacency matrix, and I don't it will be easy to define topological minors using that.Deedlit11 (talk) 12:49, March 18, 2013 (UTC) extending the function? Is it possible to generalize or extend the SCG function to make it more powerful? Just curious. --Ixfd64 (talk) 05:17, April 17, 2013 (UTC) There is certainly no obvious way to escape corresponding ordinal of SCG function. Giving more freedom for size of graphs won't increase asymptotic growth, and throwing subcubicity away may break totality of function (infinite sequences). We can allow general simple graphs if we consider graph (instead of topological) minor relation, which is WQO over set of all graphs. But it still doesn't have larger ordinal. LittlePeng9 (talk) 05:31, April 17, 2013 (UTC) I'll use Minor(n) for Robertson-Seymour function. While SCG(n) and Minor(n) functions have the same minimum ordinal, neither has a good maximum ordinal (except perhaps the proof-theoretic ordinal of a much stronger theory), and it's quite possible that the ordinal for Minor(n) is larger than the ordinal for SCG(n). Deedlit11 (talk) 06:49, April 17, 2013 (UTC) :What is the function Minor(n)? Where I can read about it? Konkhra (talk) 12:51, May 24, 2013 (UTC) ::Wonder how large Minor(1) or Minor(2)? Konkhra (talk) 03:14, May 25, 2013 (UTC) ::Minor(n) is length of longest sequence of simple graphs G such that Gi has at most i+n vertices and no Gi is graph minor of Gj for i=SSCG(n), I also guess Minor(2)=SSCG(2). LittlePeng9 (talk) 08:28, May 25, 2013 (UTC)